3.138 \(\int x (d+c^2 d x^2)^{5/2} (a+b \sinh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=193 \[ \frac{\left (c^2 d x^2+d\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 c^2 d}-\frac{b c^5 d^2 x^7 \sqrt{c^2 d x^2+d}}{49 \sqrt{c^2 x^2+1}}-\frac{3 b c^3 d^2 x^5 \sqrt{c^2 d x^2+d}}{35 \sqrt{c^2 x^2+1}}-\frac{b c d^2 x^3 \sqrt{c^2 d x^2+d}}{7 \sqrt{c^2 x^2+1}}-\frac{b d^2 x \sqrt{c^2 d x^2+d}}{7 c \sqrt{c^2 x^2+1}} \]

[Out]

-(b*d^2*x*Sqrt[d + c^2*d*x^2])/(7*c*Sqrt[1 + c^2*x^2]) - (b*c*d^2*x^3*Sqrt[d + c^2*d*x^2])/(7*Sqrt[1 + c^2*x^2
]) - (3*b*c^3*d^2*x^5*Sqrt[d + c^2*d*x^2])/(35*Sqrt[1 + c^2*x^2]) - (b*c^5*d^2*x^7*Sqrt[d + c^2*d*x^2])/(49*Sq
rt[1 + c^2*x^2]) + ((d + c^2*d*x^2)^(7/2)*(a + b*ArcSinh[c*x]))/(7*c^2*d)

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Rubi [A]  time = 0.0881364, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {5717, 194} \[ \frac{\left (c^2 d x^2+d\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 c^2 d}-\frac{b c^5 d^2 x^7 \sqrt{c^2 d x^2+d}}{49 \sqrt{c^2 x^2+1}}-\frac{3 b c^3 d^2 x^5 \sqrt{c^2 d x^2+d}}{35 \sqrt{c^2 x^2+1}}-\frac{b c d^2 x^3 \sqrt{c^2 d x^2+d}}{7 \sqrt{c^2 x^2+1}}-\frac{b d^2 x \sqrt{c^2 d x^2+d}}{7 c \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[x*(d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x]),x]

[Out]

-(b*d^2*x*Sqrt[d + c^2*d*x^2])/(7*c*Sqrt[1 + c^2*x^2]) - (b*c*d^2*x^3*Sqrt[d + c^2*d*x^2])/(7*Sqrt[1 + c^2*x^2
]) - (3*b*c^3*d^2*x^5*Sqrt[d + c^2*d*x^2])/(35*Sqrt[1 + c^2*x^2]) - (b*c^5*d^2*x^7*Sqrt[d + c^2*d*x^2])/(49*Sq
rt[1 + c^2*x^2]) + ((d + c^2*d*x^2)^(7/2)*(a + b*ArcSinh[c*x]))/(7*c^2*d)

Rule 5717

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)
^(p + 1)*(a + b*ArcSinh[c*x])^n)/(2*e*(p + 1)), x] - Dist[(b*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p +
1)*(1 + c^2*x^2)^FracPart[p]), Int[(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x] /; FreeQ[{a,
b, c, d, e, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && NeQ[p, -1]

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int x \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac{\left (d+c^2 d x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 c^2 d}-\frac{\left (b d^2 \sqrt{d+c^2 d x^2}\right ) \int \left (1+c^2 x^2\right )^3 \, dx}{7 c \sqrt{1+c^2 x^2}}\\ &=\frac{\left (d+c^2 d x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 c^2 d}-\frac{\left (b d^2 \sqrt{d+c^2 d x^2}\right ) \int \left (1+3 c^2 x^2+3 c^4 x^4+c^6 x^6\right ) \, dx}{7 c \sqrt{1+c^2 x^2}}\\ &=-\frac{b d^2 x \sqrt{d+c^2 d x^2}}{7 c \sqrt{1+c^2 x^2}}-\frac{b c d^2 x^3 \sqrt{d+c^2 d x^2}}{7 \sqrt{1+c^2 x^2}}-\frac{3 b c^3 d^2 x^5 \sqrt{d+c^2 d x^2}}{35 \sqrt{1+c^2 x^2}}-\frac{b c^5 d^2 x^7 \sqrt{d+c^2 d x^2}}{49 \sqrt{1+c^2 x^2}}+\frac{\left (d+c^2 d x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 c^2 d}\\ \end{align*}

Mathematica [A]  time = 0.147508, size = 112, normalized size = 0.58 \[ \frac{d^2 \sqrt{c^2 d x^2+d} \left (35 a \left (c^2 x^2+1\right )^4-b c x \left (5 c^6 x^6+21 c^4 x^4+35 c^2 x^2+35\right ) \sqrt{c^2 x^2+1}+35 b \left (c^2 x^2+1\right )^4 \sinh ^{-1}(c x)\right )}{245 c^2 \left (c^2 x^2+1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x]),x]

[Out]

(d^2*Sqrt[d + c^2*d*x^2]*(35*a*(1 + c^2*x^2)^4 - b*c*x*Sqrt[1 + c^2*x^2]*(35 + 35*c^2*x^2 + 21*c^4*x^4 + 5*c^6
*x^6) + 35*b*(1 + c^2*x^2)^4*ArcSinh[c*x]))/(245*c^2*(1 + c^2*x^2))

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Maple [B]  time = 0.18, size = 863, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x)),x)

[Out]

1/7*a/c^2/d*(c^2*d*x^2+d)^(7/2)+b*(1/6272*(d*(c^2*x^2+1))^(1/2)*(64*c^8*x^8+64*c^7*x^7*(c^2*x^2+1)^(1/2)+144*c
^6*x^6+112*c^5*x^5*(c^2*x^2+1)^(1/2)+104*c^4*x^4+56*c^3*x^3*(c^2*x^2+1)^(1/2)+25*c^2*x^2+7*c*x*(c^2*x^2+1)^(1/
2)+1)*(-1+7*arcsinh(c*x))*d^2/c^2/(c^2*x^2+1)+1/640*(d*(c^2*x^2+1))^(1/2)*(16*c^6*x^6+16*c^5*x^5*(c^2*x^2+1)^(
1/2)+28*c^4*x^4+20*c^3*x^3*(c^2*x^2+1)^(1/2)+13*c^2*x^2+5*c*x*(c^2*x^2+1)^(1/2)+1)*(-1+5*arcsinh(c*x))*d^2/c^2
/(c^2*x^2+1)+1/128*(d*(c^2*x^2+1))^(1/2)*(4*c^4*x^4+4*c^3*x^3*(c^2*x^2+1)^(1/2)+5*c^2*x^2+3*c*x*(c^2*x^2+1)^(1
/2)+1)*(-1+3*arcsinh(c*x))*d^2/c^2/(c^2*x^2+1)+5/128*(d*(c^2*x^2+1))^(1/2)*(c^2*x^2+c*x*(c^2*x^2+1)^(1/2)+1)*(
-1+arcsinh(c*x))*d^2/c^2/(c^2*x^2+1)+5/128*(d*(c^2*x^2+1))^(1/2)*(c^2*x^2-c*x*(c^2*x^2+1)^(1/2)+1)*(1+arcsinh(
c*x))*d^2/c^2/(c^2*x^2+1)+1/128*(d*(c^2*x^2+1))^(1/2)*(4*c^4*x^4-4*c^3*x^3*(c^2*x^2+1)^(1/2)+5*c^2*x^2-3*c*x*(
c^2*x^2+1)^(1/2)+1)*(1+3*arcsinh(c*x))*d^2/c^2/(c^2*x^2+1)+1/640*(d*(c^2*x^2+1))^(1/2)*(16*c^6*x^6-16*c^5*x^5*
(c^2*x^2+1)^(1/2)+28*c^4*x^4-20*c^3*x^3*(c^2*x^2+1)^(1/2)+13*c^2*x^2-5*c*x*(c^2*x^2+1)^(1/2)+1)*(1+5*arcsinh(c
*x))*d^2/c^2/(c^2*x^2+1)+1/6272*(d*(c^2*x^2+1))^(1/2)*(64*c^8*x^8-64*c^7*x^7*(c^2*x^2+1)^(1/2)+144*c^6*x^6-112
*c^5*x^5*(c^2*x^2+1)^(1/2)+104*c^4*x^4-56*c^3*x^3*(c^2*x^2+1)^(1/2)+25*c^2*x^2-7*c*x*(c^2*x^2+1)^(1/2)+1)*(1+7
*arcsinh(c*x))*d^2/c^2/(c^2*x^2+1))

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Maxima [A]  time = 1.14682, size = 130, normalized size = 0.67 \begin{align*} \frac{{\left (c^{2} d x^{2} + d\right )}^{\frac{7}{2}} b \operatorname{arsinh}\left (c x\right )}{7 \, c^{2} d} + \frac{{\left (c^{2} d x^{2} + d\right )}^{\frac{7}{2}} a}{7 \, c^{2} d} - \frac{{\left (5 \, c^{6} d^{\frac{7}{2}} x^{7} + 21 \, c^{4} d^{\frac{7}{2}} x^{5} + 35 \, c^{2} d^{\frac{7}{2}} x^{3} + 35 \, d^{\frac{7}{2}} x\right )} b}{245 \, c d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x)),x, algorithm="maxima")

[Out]

1/7*(c^2*d*x^2 + d)^(7/2)*b*arcsinh(c*x)/(c^2*d) + 1/7*(c^2*d*x^2 + d)^(7/2)*a/(c^2*d) - 1/245*(5*c^6*d^(7/2)*
x^7 + 21*c^4*d^(7/2)*x^5 + 35*c^2*d^(7/2)*x^3 + 35*d^(7/2)*x)*b/(c*d)

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Fricas [A]  time = 2.61141, size = 483, normalized size = 2.5 \begin{align*} \frac{35 \,{\left (b c^{8} d^{2} x^{8} + 4 \, b c^{6} d^{2} x^{6} + 6 \, b c^{4} d^{2} x^{4} + 4 \, b c^{2} d^{2} x^{2} + b d^{2}\right )} \sqrt{c^{2} d x^{2} + d} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) +{\left (35 \, a c^{8} d^{2} x^{8} + 140 \, a c^{6} d^{2} x^{6} + 210 \, a c^{4} d^{2} x^{4} + 140 \, a c^{2} d^{2} x^{2} + 35 \, a d^{2} -{\left (5 \, b c^{7} d^{2} x^{7} + 21 \, b c^{5} d^{2} x^{5} + 35 \, b c^{3} d^{2} x^{3} + 35 \, b c d^{2} x\right )} \sqrt{c^{2} x^{2} + 1}\right )} \sqrt{c^{2} d x^{2} + d}}{245 \,{\left (c^{4} x^{2} + c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x)),x, algorithm="fricas")

[Out]

1/245*(35*(b*c^8*d^2*x^8 + 4*b*c^6*d^2*x^6 + 6*b*c^4*d^2*x^4 + 4*b*c^2*d^2*x^2 + b*d^2)*sqrt(c^2*d*x^2 + d)*lo
g(c*x + sqrt(c^2*x^2 + 1)) + (35*a*c^8*d^2*x^8 + 140*a*c^6*d^2*x^6 + 210*a*c^4*d^2*x^4 + 140*a*c^2*d^2*x^2 + 3
5*a*d^2 - (5*b*c^7*d^2*x^7 + 21*b*c^5*d^2*x^5 + 35*b*c^3*d^2*x^3 + 35*b*c*d^2*x)*sqrt(c^2*x^2 + 1))*sqrt(c^2*d
*x^2 + d))/(c^4*x^2 + c^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c**2*d*x**2+d)**(5/2)*(a+b*asinh(c*x)),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x)),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError